Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
n = int(raw_input())
p = map(float, raw_input().split())
dp = [[0.00 for i in xrange(3001)] for _ in xrange(3001)]

dp[0][1] = p[0]
dp[0][0] = 1.0-p[0]

for i in xrange(1, n):
    for j in xrange(i + 1):
        dp[i][j] += (1.0 - p[i]) * dp[i-1][j]
        dp[i][j+1] += p[i] * dp[i-1][j]

res = 0.0
for i in xrange(n + 1):
    red = n - i
    if i > red:
        res += dp[n-1][i]
print res